Let S be the portion of the paraboloid: z=4-x^2-y^2 that lays above the xy-axis.

Question

Let S be the portion of the paraboloid: z=4-x^2-y^2 that lays above the xy-axis. Then r(u,v)=(ucosv, usinv, 4-u^2) where 0<=u<=2, 0<=v<=2pie defines the surface. Suppose F(x,y,z)= (x,y,z). Find the flux out at S.

Explanation / Answer

Below is the correct procedure. please follow the method and find the answer .ThanQ Projecting into the xy-plane, the region is bounded in the circle x^2 + y^2 = 4 ==> The bounds are y = -sqrt(4 - x^2) to sqrt(4 - x^2), with x = -2 to 2. Moreover, the unit normal is given by (-z_x, -z_y, 1) = (-2x, -2y, 1). Hence, ??s F · dS = ?(x = -2 to 2) ?(y = -sqrt(4 - x^2) to sqrt(4 - x^2)) (0, x, x^2 + y^2) · (-2x, -2y, 1) dy dx = ?(x = -2 to 2) ?(y = -sqrt(4 - x^2) to sqrt(4 - x^2)) (x^2 - 2xy + y^2) dy dx = ?(x = -2 to 2) [x^2 * 2 sqrt(4 - x^2) - 0 + 2 (4 - x^2)^(3/2) / 3] dx = 2 ?(x = 0 to 2) [2x^2 (4 - x^2)^(1/2) + (2/3) (4 - x^2)^(3/2)] dx, since the integrand is even = (4/3) ?(x = 0 to 2) [3x^2 (4 - x^2)^(1/2) + (4 - x^2)^(3/2)] dx Now, let x = 2 sin t, dx = 2 cos t dt. Bounds: x = 0 ==> t = 0 and x = 2 ==> t = p/2. ==> (4/3) ?(t = 0 to p/2) [3 (4 sin^2(t)) (2 cos t) + 8 cos^3(t)] (2 cos t dt) = (64/3) ?(t = 0 to p/2) [3 sin^2(t) cos^2(t) + cos^2(t) cos^2(t)] dt = (64/3) ?(t = 0 to p/2) [3 sin^2(t) + cos^2(t)] cos^2(t) dt = (64/3) ?(t = 0 to p/2) [2 sin^2(t) + 1] cos^2(t) dt = (64/3) ?(t = 0 to p/2) [(1 - cos(2t)) + 1] (1/2)(1 + cos(2t)) dt = (32/3) ?(t = 0 to p/2) (2 + cos(2t) - cos^2(2t)) dt = (32/3) ?(t = 0 to p/2) [2 + cos(2t) - (1/2)(1 + cos(4t))] dt = (16/3) ?(t = 0 to p/2) [3 + 2 cos(2t) - cos(4t)] dt = (16/3) [3t + sin(2t) + sin(4t)/4] {for t = 0 to p/2} = 8p. I hope this helps!

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